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3 years ago

Hello!! I need help please! Please explain how I can identify a percipitate in a reaction! I need help with chemistry ha

1 answer:

5 0

Generally speaking a precipitate is something that will form as a result of the reaction. It generally is a solid material.

Sodium and nitrates are soluble so Calcium Phosphate would precipitate out....

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Which molecule is pentanoic acid?

Lana71 [14]

Answer:

Explanation:carboxyli

7 0

3 years ago

The melting point of H₂O(s) is 0 °C. Would you expect the melting point of H₂S(s) to be 85 °C, 0 °C or -85 °C.? Justify your cho

dimulka [17.4K]

Answer:

-85 °C

Explanation:

O and S are in the same group( Group 16). Since S is below O it's atomic mass is higher than O. So molar mass of H2S is higher than H2O. The strength of Vanderwaal Interactions ( London dispersion forces) increases when the molar mass increases. However, only H2O can form H bonds with each other. This is because electronegativity of O is higher than S and therefore H in H2O has a higher partial positive charge than H of H2S.

H bond dominate among these 2 types of forces so the strength of attractions between molecules is higher in H2O than H2S. Therefore more energy should be supplied for H2O to break inter

molecular forces and convert from solid to liquid state than H2S. So mpt of H2O must be higher than that of H2S.

5 0

3 years ago

How many grams in 82 moles of CO2

Tems11 [23]

44.0 g/mol

hope it helps

6 0

3 years ago

Consider the following chemical reaction: 2KCl + 3O2 --> 2KClO3. If you are given 100.0 moles of KCl and 100.0 moles of O2...

g100num [7]

Answer:

O₂; KCl; 33.3

Explanation:

We are given the moles of two reactants, so this is a limiting reactant problem.

We know that we will need moles, so, lets assemble all the data in one place.

2KCl + 3O₂ ⟶ 2KClO₃

n/mol: 100.0 100.0

1. Identify the limiting reactant

(a) Calculate the moles of KClO₃ that can be formed from each reactant

(i)From KCl

$\text{Moles of KClO}_{3} = \text{100.0 mol KCl} \times \dfrac{\text{2 mol KClO}_{3}}{\text{2 mol KCl}} = \text{100.0 mol KClO}_{3}$

(ii) From O₂

$\text{Moles of KClO}_{3} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KClO}_{3}}{\text{3 mol O}_{2}} = \text{66.67 mol KClO}_{3}$

O₂ is the limiting reactant, because it forms fewer moles of the KClO₃.

KClO₃ is the excess reactant.

2. Moles of KCl left over

(a) Moles of KCl used

$\text{Moles used} = \text{100.0 mol O}_{2} \times \dfrac{\text{2 mol KCl}}{\text{3 mol O}_{2}} = \text{66.67 mol KCl}$

(b) Moles of KCl left over

n = 100.0 mol - 66.67 mol = 33.3 mol

3 0

3 years ago

Match the following with their correct molecular weight. 2-butanone Propyl acetate 4-methyl-2-pentanone Butyl acetate Methanol E

Hatshy [7]

Answer:

2-butanone = 72.11 g/mol (option F)

Propyl acetate = 102.13 g/mol (option C)

4-methyl-2-pentanone = 100.16 g/mol (option D)

Butyl acetate = 116.16 g/mol (option B)

Methanol = 32.04 g/mol (option E)

Ethanol = 46.07 g/mol (option A)

Explanation:

Step 1: Data given

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2:

2-butanone = C4H8O

⇒ 4*12.01 + 8*1.01 + 16.00 = 72.11 g/mol (option F)

Propyl acetate = C5H10O2

⇒ 5*12.01 + 10*1.01 + 2*16.00 = 102.13 g/mol (option C)

4-methyl-2-pentanone = C6H12O

⇒ 6*12.01 + 12*1.01 + 16.00 = 100.16 g/mol (option D)

Butyl acetate = C6H12O2

⇒ 6*12.01 + 12*1.01 + 2*16.00 = 116.16 g/mol (option B)

Methanol = CH3OH = CH4O

⇒ 12.01 + 4*1.01 + 16.00 = 32.04 g/mol (option E)

Ethanol = C2H5OH = C2H6O

⇒ 2*12.01 + 6*1.01 + 16.00 = 46.07 g/mol (option A)

4 0

3 years ago